Pembahasan Paket Soal

Oleh Fatimah Tuzzahroh

9 tayangan
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Transkrip Pembahasan Paket Soal

Pembahasan Paket Soal
1. Penyelesaian:
BC = √ A B 2−A C 2
= √ 172−152
= √ 289−225
= √ 64=8
Jawaban: D

2. Penyelesaian:
∠ PRQ = 90°−¿ 60° =30°
1
cos30°= √ 3
2
PR
cos PRQ =
QR
√3 = PR
2 18
PR=9 √ 3
Jawaban: C
3. Penyelesaian:
5
Sinα =
9
R
5

9

α

P

BC =
=
=
=

Q

√ A C 2− A B 2
√ 92−5 2

√ 81−25
√ 56= √ 4 ×14=2 √14
BC 2 √ 14
=
Cosα =
AC
Jawaban: B

9

4. Penyelesaian:
13
5
Secα  cosα
5
13
BC = √ A B 2−A C 2
= √ 13²−52
= √ 169−25
= √ 144=12
BC 12
=
sinα =
AB 13
jawaban: D

R
13

α

P

5

Q

5. Penyelesaian:
1
1
sinA−¿sinB = 2 cos ( A+ B ) .sin ( A−B)
2
2
maka:
1
1
sin75°−¿sin165° = 2 cos (240 ° ) . sin (−90°)
2
2
= 2 cos 120 ° .−sin 45 °
−1 −1
1
√2 = √2
=2
2
2
2
Jawaban: D

( )(

6. Penyelesaian:
tan75° = tan(45° + 30°)
tan 45 ° + tan30 °
=
1−tan 45° . tan 30 °
3+ √ 3
3
.
=
3
3−√ 3
3+ √ 3 3+ √ 3
.
=
3−√ 3 3+ √ 3
9+6 √ 3+3
=
9−3
12+ 6 √3
=2+ √ 3
=
6
Jawaban: B

)

7. Penyelesaian:
1
tanx =
2
3tanx −tan3 x
tan3x =
1−3 tan ² x
1
1
3

³
2
2
=
1
1−3
²
2
3 1 12−1 11

2 8
8
8 11 4 11
1
=
= = × = =5
=
3
4−3
1
8 1 2
2
1−
4
4
4
Jawaban: A

()()
()

8. Penyelesaian:
( √ 2 , √ 6 )x=√ 2 dan y=√ 6
r= √ x ²+ y ²
= √ ( √2 ) ²+ ( √ 6 ) ²= √2+6=√ 8=2 √ 2

y √ 6 √ 2 √ 12
= × =
= √3
x √2 √ 2
2
tanθ bernilai √ 3 untuk θ =60°
jadi, koordinat kutubnya adalah (2√ 2,60°)
9. Penyelesaian:
( cosx+ sinx )2
cos ² x +sin ²+ 2 sinx . cosx
2 =
cos
² x +sin ² x−2 sinx . cosx
( cosx−sinx )

tanθ =

=

1+2 sinx . cosx 1+sin 2 x
=
1−2 sinx . cosx 1−sin 2 x

Jawaban D
10. Penyelesaian:
∠ A siku-siku= 90°
C

A

B

cos (A+B) = p
cos (A+B) = cos (90°+B)
= −¿sinB
−¿sinB = p

SinB = −¿ p
B+C = 90°
cosC = cos(90°−¿B)= sinB
sinB + cosC = sinB + sinB
= 2 sin B
= −¿ 2p
Jawaban: E
11. Penyelesaian:
tan²x −¿1 = p³
2
sin x
−1 = p³
co s2 x
2
sin x
= p³ + 1
2
co s x
1−co s2 x
= p³ + 1
2
co s x
1−cos ² x = p³ cos²x + cos²x
1 = p³ cos²x + cos²x + cos²x
1 = p³ cos²x + 2cos²x
1 = (p³ + 2) cos²x
1
cos²x= 3
p +2
Jawaban: C

12. Penyelesaian:
P

p+r= 18
60°

∠ Q= 30°
30°

R

 cos30° =

∠ P= 60°
Q

QR
PQ

√3 = p

r
2
2p = √ 3 r
√3 r
p=
2
 p + r = 18
√3 r + r = 18
2
√ 3+ 2 r=18
2
2
36
r =18×
=
2+ √ 3 2+ √ 3
PR
 sin 30° =
PQ
1 PR
=
2
r
36
2PR =
2+ √ 3
36
1
×
PR =
2+ √3 2
18
2−√ 3
×
PR =
2+ √3 2−√ 3
36−18 √ 3
=
= 36 - 18 √ 3
4−3
Jawaban: D

(

)

13. penyelesaian:

=

=

sin 220 °−sin 80 °
cos 220 °−cos 80 °
220 °+ 80 °
220 °+ 80 °
2cos
. sin
2
2

(
) (
)
220 °+ 80°
220° + 80°
−2sin (
.sin (
)
)
2
2

2cos 150 ° . sin70 °
−2sin 50 ° .sin 70 °
2cos 150 °
=
−2sin 50 °
−1
2
√3
2
−√ 3
=
=
= √3
−1
1
−2
2
Jawaban: A

=

(

()

)

14. Peyelesaian:
cos2α = 2cos²α −1

(5)
6
= 2( )−1
25
2

=2

√ 6 −1

12
12−25 13
−1=
=
25
25
25
Jawaban: B
15. Penyelesaian:
2
cosA.cosB =
3
1
2
(cos(A+B)+cos(A-B)) =
2
3
4
(cos(A+B)+cos(A-B)) =
3
π
4
cos(A+B)+cos( ) =
6
3
4
cos(A+B)+cos(30°) =
3

=

√3 = 4

cos(A+B)+

2

3

4 √3
3 2
8− √ 3
cos(A+B) =
6
Jawaban: C
16. Penyelesaian:
QA= √ P A 2−PQ 2

cos(A+B) =

P

= √ 13²−52
= √ 169−25
= √ 144=12

13

5

Tan A =
A

Q

BL = √ B K 2−K L2

= √ 17²−8 2
= √ 289−64
17 = √ 225=15

K

8

Tan B =
B

L

Tan(A-B) =

5
12

8
12

tan A−tan B
1−tanA .tanB

5 8
75−96

12 15
180
−21 180 −21
=
=
×
=
=
5 8
180−40 180 140 140
1− .
12 15
180
Jawaban: E
17. Penyelesaian:
QR
PQ
=
sinP sinR

QR
PQ
=
sin 30° sin 45 °
QR
12
1 = 1
√2
2
2
QR√ 2=12
12 √2
×
QR =
√2 √2
12 √ 2
=
√2
= 6√ 2
Jawaban: D
18. Penyelesaian:
PR = 8 cm, PQ = 5 cm, dan ∠ RPQ= 60°
Q
r
5 cm

p
R
60°

8 cm

q

P

p² = q²+r²-2q.r.cosP
= 8²+5²-2.8.5.cos 60°
1
= 64 + 25 – 80
2
= 60 + 25 – 40
= 45
p = √ 45 cm
Jawaban: A
19. Penyelesaian:
Cos2x – 3 cosx + 2= 0
2 cos²x – 1 – 3 cosx + 2= 0
2 cos²x – 3 cosx + 1= 0
(2cosx – 1)(cosx – 1)=0
2 cosx – 1 =0
2 cos x =1

()

Cosx =

1
2

 2 cosx – 1 =0
2 cos x = 1
1
Cosx =
2
π 5
, π
x=
3 3
 Cosx – 1 = 0
Cosx = 1
x= {0,2 π }
π 5
HP = 0 , , π , 2 π
3 3
Jawaban: B
20. Penyelesaian:
1
cosx°=
2
cosx°= cos 60°
x°= ±60 + k.360°
*x= 60°+k.360°
x= 60°
*x= -60°+ k.360°
x= 300°
Jadi, nilai x yang memenuhi adalah 60° dan 300°.
Jawaban: C
21. Penyelesaian:
Sin60°(sin60°-x) = sin60°cosx – cos60°.sinx
1
1
= √ 3 cosx− sinx
2
2
−1
1
sinx + √ 3 cosx
=
2
2
1
1
sinx + √ 3cosx = p sinx + q cosx
2
2
1
p¿
2
1
q= √ 3
2

{

}

{

}

( )

−1 1
+ √ 3 ( √3 )
2
2
−1 3 2
+ = =1
=
2 2 2
Jawaban: B

p+q√ 3=

Judul: Pembahasan Paket Soal

Oleh: Fatimah Tuzzahroh


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